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I am currently going through Introduction to Electrodynamics by Griffiths. 4th ed. In the book p.16 problem 1.14, I noticed an expression like this:

For $f(y,z)$ and $\bar{y}(y,z)$, $$\frac{\partial f}{\partial \bar{y}}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial \bar{y}} +\frac{\partial f}{\partial z}\frac{\partial z}{\partial \bar{y}}.$$

And the book, of course, uses $\partial$ for partial derivative as well. I am confused about the symbol. Is it a convention of the total derivative in EM or is it something different? I hate picking on insignificant details, but the fear is I've forgotten more about vector calculus than I anticipated.

So the question is

Suppose that f is a function of two variables (y and z) only. Show that the gradient $\nabla f = (\frac{∂f}{∂y}\hat{y} + \frac{∂f}{∂z}\hat{z})$ transforms as a vector under rotations.

This is not about formalism, I was silly to be distracted by it. I figure I will post what I found out as the answer. Since the answer is the two answer combined.

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@Rururu, if you read the entire question as given in the book, $f=f(y,z)$ and we have $\bar{y}=\bar{y}(y,z)$ and $\bar{z}=\bar{z}(y,z)$ from which we can compute $y=y(\bar{y},\bar{z})$ and $z=z(\bar{y},\bar{z})$. As a result, $f=f(y(\bar{y},\bar{z}),z(\bar{y},\bar{z}))$ i.e. $f=f(\bar{y},\bar{z})$. So, as a result, $\frac{\partial f}{\partial \bar{y}}$ is valid and the use of $\partial$ is also valid and the expressed expansion is just the application of chain rule of multivariable calculus. See this example .

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  • $\begingroup$ I understand what you are saying but to my understanding the process is noted as $\frac{df}{d\bar {y}}$ instead of $\frac{\partial f}{\partial \bar{y}}$. Can you clarify on this? Thank you in advance. $\endgroup$ – Rururu Jun 26 '18 at 9:43
  • $\begingroup$ Do you agree that $f$ is a multivariable function wrt $y,z$ right? $\endgroup$ – Naman Agarwal Jun 26 '18 at 9:47
  • $\begingroup$ If so, switching to $\bar{y}$ and $\bar{z}$ is just a change of variable and so, f will be a multivariable function of $\bar{y}$ and $\bar{z}$ and so, variation of $f$ wrt $\bar{y}$ is partial keeping $\bar{z}$ constant. Checkout the link to the example I gave in the answer. It would help you understand better. $\endgroup$ – Naman Agarwal Jun 26 '18 at 9:49
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See this pdf: https://www.icp.uni-stuttgart.de/~icp/mediawiki/images/7/74/Remark_on_partial_derivatives.pdf

I think the pdf should make clear that in physics the partial derivative is used in a nasty way.

In principle the pdf shows that in physics partial derivatives are often used to act on abbreviations. Although they shall not act on these abbreviations if you follow the mathematical definition in a strict sense. E.g. in statistical mechanics beta=1/(kT) is introduced as abbreviation. If you calculate $$c_V=\frac{\partial }{\partial T} \frac{\sum_E \Omega(E) E \exp(-\beta E)}{\sum_E \Omega(E) \exp(-\beta E)},$$ it would yield zero if you follow the definition of the partial derivative in a strict way. Since the term only depends on $\beta$. However a physicist will realize that $\beta$ was only introduced as abbreviation and will use the chain rule to calculate: $$c_V=\frac{\partial }{\partial \beta} \frac{\sum_E \Omega(E) E \exp(-\beta E)}{\sum_E \Omega(E) \exp(-\beta E)} \frac{\partial \beta}{\partial T}.$$ While this is not the standard usage of the partial derivative it is common to use under physicists. Since this is in principle undefined behaviour it is important that the author notes somewhere that he wants to act the $\partial$ sign in that way. Note that the chain rule cannot be applied here unless it is made clear by the author that the sign $\partial$ is meant to act on $\beta$ as well! If the author forgets to make this clear his result is (from a strict mathematical point of view) wrong.

See the following pdf for more details!

https://www.icp.uni-stuttgart.de/~icp/mediawiki/images/7/74/Remark_on_partial_derivatives.pdf

Having said all this and coming back to your question: the author in the book just introduced $\overline{y}$ as an abbreviation and therefore employs the chain rule here. This is however (to repeat myself) not the strict mathematical definition of the partial derivative. Knowing that he uses $\overline{y}$ as abbreviation you shuould howver be able to reproduce all his formulas using what Naman Agarwal suggested to you.

Please note some formal problems with Naman Agarwals answer: $f(x,y)=f(\overline{x},\overline{y})$ is wrong in general. He should introduce a new function $\tilde f$. It should read $$f(x,y)=\tilde{f}(\overline{x},\overline{y}).$$ Imagine $f(x,y)=sin(x)\cdot y$. Now using $\overline{x}=x-3$ and $\overline{y}=x/y$, we would have $$f(\overline{x},\overline{y})=\sin(x-3)\cdot\frac{x}{y}$$ which is not equal to $$f(x,y)=\sin(x)\cdot y.$$ Clearly a new function $$\tilde{f}(a,b)=\sin(a+3)\cdot\frac{a+3}{b}$$ is needed such that $$f(x,y)=\tilde{f}(\overline{x},\overline{y}).$$

Note that also the example on Wikipedia is strictly speaking wrong: $$\frac{\partial u(x(r),y(r))}{\partial r}=0$$ is true. No chain rule shall be applied if $\partial$ is the truly mathematically defined partial derivative.

Consider the total derivative: $$\frac{d u(x(r),y(r),0r)}{d r}=\frac{\partial u(x(r),y(r))}{\partial x} \frac{\partial x}{\partial r}+ \frac{\partial u(x(r),y(r))}{\partial y} \frac{\partial y}{\partial r}+ \frac{\partial u(x(r),y(r))}{\partial r} $$ You see that $$\frac{\partial u(x(r),y(r))}{\partial r}=0$$

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